LeetCode1005☞K次取反后最大的数组和

news2025/3/18 3:25:34

关联LeetCode题号1005

本题特点

贪心：局部最优解：将负数取反得到比原值大的值，进而全局最优解整体和为最大
二次贪心: 如果取反次数大于负数个数，那么剩下次数如果为奇数，那么就将绝对值最小的数取反（贪心贪在：正数绝对值取反得到的值最小局部最优，进而全局最优得到最大的和）

本题思路

情况1：按照从大到小的排序，如果k不为0并且负数进行取反，
情况2：若k有剩余，并且剩余未奇数，将绝对值最小的数取反，偶数则无影响

class Solution:
    def largestSumAfterKNegations(self, nums: List[int], k: int) -> int:
        nums.sort(key=lambda x: abs(x), reverse=True)

        for i in range(len(nums)):
            if nums[i] < 0 and k != 0:
                nums[i] *= -1
                k -= 1
        if k % 2 == 1:
            nums[-1] *= -1

        return sum(nums)

Java写法

package leetcode;

import org.junit.jupiter.api.Test;

import java.util.Arrays;

import java.util.stream.IntStream;

/**
 * File Description: MaximizeSumOfArrayAfterKNegations
 * Author: Your Name
 * Date: 2024/12/23
 */
public class MaximizeSumOfArrayAfterKNegations_1005 {


    public int largestSumAfterKNegations(int[] nums, int k) {
        Arrays.sort(nums);
        for (int i=0; i < nums.length; i++){
            if (nums[i] < 0){
                nums[i] *= -1;
                k --;
            }
        }
        if(k % 2 == 1){
            Arrays.sort(nums);
            nums[0] *= -1;
        }
        int sum = 0;
        for(int n: nums){
            sum += n;
        }
        return sum;
    }

    public int largestSumAfterKNegations2(int[] nums, int k) {
        nums = IntStream.of(nums).boxed().sorted((o1, o2) -> Math.abs(o2) - Math.abs(o1)).mapToInt(Integer::intValue).toArray();
        for (int i=0; i < nums.length; i++){
            if (nums[i] < 0 && k != 0){
                nums[i] *= -1;
                k --;
            }
        }
        if (k % 2 == 1){
            nums[nums.length-1] *= -1;
        }

        int sum = IntStream.of(nums).sum();
        return sum;

    }

    @Test
    public void largestSumAfterKNegationsTest(){
        int[] nums = {3, -1, 0, 2};
        int s = largestSumAfterKNegations(nums, 3);
        int s1 = largestSumAfterKNegations2(nums, 3);
        System.out.println(s1);
    }

}